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2008 iTest Problems/Problem 74

Revision as of 17:42, 21 November 2018 by Rockmanex3 (talk | contribs) (Solution to Problem 74 -- centroids and quads)
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Problem

Points $C$ and $D$ lie on opposite sides of line $\overline{AB}$. Let $M$ and $N$ be the centroids of $\triangle ABC$ and $\triangle ABD$ respectively. If $AB=841$, $BC=840$, $AC=41$, $AD=609$, and $BD=580$, find the sum of the numerator and denominator of the value of $MN$ when expressed as a fraction in lowest terms.

Solution

Notice that $41^2 + 840^2 = 841^2$ and $580^2 + 609^2 = 841^2$, so $\triangle ABC$ and $\triangle ABD$ are right triangles. Since $\angle ACB = \angle ADB = 90^\circ$, quadrilateral $ACBD$ is cyclic.

[asy] pair A=(0,0),C=(1681/841,34440/841),B=(841,0),D=(370881/841,-353220/841); draw(A--C--B--D--A); dot(A); label("A",A,W); dot(C); label("C",C,NW); dot(B); label("B",B,E); dot(D); label("D",D,S); draw(circle((841/2,0),841/2)); dot((841/2,0)); draw(C--(841/2,0)--D); pair M=(1417924/5046,11480/841),n=(1282/3,-140); dot(M); dot(n); label("M",M,N); label("N",n,E); draw(C--D,dotted); draw(M--n,dotted); [/asy]

Let $O$ be the midpoint of $AB$. Since $M$ and $N$ are centroids, $CM : MO = DN : DO = 2 : 1$. Thus, by SAS Similarity, $\triangle COD \sim \triangle MON$, so $MN = \tfrac13 CD$.

By Ptolemy's Theorem, $AC \cdot BD + BC \cdot AD = CD \cdot AB$, so \begin{align*} 41 \cdot 580 + 840 \cdot 609 &= 841 \cdot CD \\ 535340 &= 841 \cdot CD \\ CD &= \frac{18460}{29} \end{align*}

That means $MN = \tfrac13 \cdot \tfrac{18460}{29} = \tfrac{18460}{87}$, and the sum of the numerator and denominator in lowest terms is $18460 + 87 = \boxed{18547}$.

See Also

2008 iTest (Problems)
Preceded by:
Problem 73 		Followed by:
Problem 75
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